53 Proof: By induction, one has |an+1 −an|≤k n−1|a2 −a1| for all n∈N. Thus, |an+p −an|≤|an+1 −an|+|an+2 −an+1|+···+|an+p −an+p −1| ≤(kn−1 +kn +···+kn+p−2)|a2 −a1| ≤kn−1(1+k+k2 +···+kp−1)|a2 −a1| ≤ kn−1 1−k| a2 −a1|. for all n, p ∈N. Since kn−1 →0 as n →∞(independently of p), this implies {an} is a Cauchy sequence and, hence, it is convergent. □ The conditionk <1 in the previous theorem is crucial. Consider the following example. ■ Example 2.4.1 Let an =lnn for all n∈N. Since 1< n+2 n+1 < n+1 n for all n∈Nand the natural logarithm is an increasing function, we have |an+2 −an+1| =|ln(n+2)−ln(n+1)| = ln n+2 n+1 =ln n+2 n+1 <ln n+1 n =|ln(n+1)−lnn| =|an+1 −an|. Therefore, the inequality in Definition 2.4.2 is satisfied with k =1, yet the sequence {lnn} does not converge since it is not bounded. Exercises 2.4.1 ▶Determine which of the following are Cauchy sequences. (a) an =(−1) n. (b) an =(−1) n/n. (c) an =n/(n+1). (d) an =(cosn)/n. 2.4.2 Prove that the sequence an = ncos(3n2 +2n+1) n+1 has a convergent subsequence.
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